|
| |
|
|
A133953
|
|
A second integer solution:d=2;h=1; A 4 X 4 vector Markov of a game matrix MA and an anti- game matrix MB such that game_valueMa+game_ValueMB =0 and the score is the sum of the vector out put of the Markov: MA={{0,1},{1,d}}; MB={{1/h,0},(2 - d + 1/h + h),h}}; Characteristic Polynomial is: -1 + 4 x^2 - 4 x^3 + x^4.
|
|
0
|
|
|
|
2, 6, 12, 24, 50, 110, 252, 592, 1410, 3382, 8140, 19624, 47346, 114270, 275836, 665888, 1607554, 3880934, 9369356, 22619576, 54608434, 131836366, 318281084, 768398448, 1855077890, 4478554134, 10812186060, 26102926152, 63018038258
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Using the general game value function: gv[M_] = Det[M]/Sum[Sum[M[[i, j]]*(-1)^(i + j), {i, 1, 2}], {j, 1, 2}] and matrices: MA = {{a, b}, {c, d}}; MB = {{e, f}, {g, h}}; The solution of: Solve[{b == 1, f == 0, a == 0, Det[MA] + 1 == 0, Det[MB] - 1 == 0, gv[MA] + gv[MB] == 0}, {c, d, e, g, h}] Gives the two Matrix solution: MA={{0,1},{1,d}}; MB={{1/h,0},{2-d+1/h+h,h}}; Besides the Fibonacci:d=1 and h=1 the only other obvious integer solution is: d=2 and h=1.
|
|
|
LINKS
|
Table of n, a(n) for n=1..29.
|
|
|
FORMULA
|
d = 2; h = 1; M = {{0, 1, 0, 0}, {1,d, 0, 0}, {0, 0, 1/h, 0}, {0, 0, 2 - d + 1/h + h, h}}; v[1] = {0, 1, 1, 0}; v[n_] := v[n] = M.v[n - 1] a(n) = Sum[v(i),{i,1,4}].
G.f.: 2*x*(1+x)*(1-2*x)/((1-x)^2*(1-2*x-x^2)). [Colin Barker, Feb 28 2012]
|
|
|
MATHEMATICA
|
d = 2; h = 1; M = {{0, 1, 0, 0}, {1, d, 0, 0}, {0, 0, 1/h, 0}, {0, 0, 2 - d + 1/h + h, h}}; v[1] = {0, 1, 1, 0}; v[n_] := v[n] = M.v[n - 1] a = Table[Apply[Plus, v[n]], {n, 1, 50}]
|
|
|
CROSSREFS
|
Sequence in context: A309841 A132176 A197469 * A122863 A170935 A277173
Adjacent sequences: A133950 A133951 A133952 * A133954 A133955 A133956
|
|
|
KEYWORD
|
nonn,uned
|
|
|
AUTHOR
|
Roger L. Bagula, Jan 08 2008
|
|
|
STATUS
|
approved
|
| |
|
|
